I've been doing some more concrete arithmetic with one of these outer automorphisms and it's working out just the way "they" say it will (e.g. $(123)(45)(6)$ goes to $(ABCDEF)$ and $(123456)$ goes to $(A)(BC)(DEF)$, etc. etc.).

My next question: If I'm not mistaken, $S_6$ should be a subgroup of some larger group of permutations, but smaller than the group of all permutations on that larger set---I would think a group of permutations on a set that has $\{1,2,3,4,5,6\}$ as a subset---such that some inner automorphism of that larger group, when restricted to $S_6$, is an outer automorphism of $S_6$. How big is that larger set and what group of permutations on it do we need to look at, and which of its members do we conjugate by?

Interesting. I'm wondering if in general given some outer automorphism of a group $G$, could you embed $G$ as a normal subgroup of some group $H$, where conjugation by an element of $H$ gives the outer automorphism? For example outer automorphisms of $A_n$ are conjugation by elements of $S_n$ when $n \neq 2, 3, 6$.
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Mikko KorhonenMar 25 '12 at 21:15

I believe that does hold generally.
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Michael HardyMar 25 '12 at 21:16

2

Yes. Just take the semidirect product of $S_6$ with $\mathrm{Aut}(S_6)$ :)
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Bill CookMar 25 '12 at 21:24

1

For "concrete arithmetic" with groups, you might consider GAP. You can easily compute the outer automorphism of $S_6$ explicitly, as explained by Alexander Hulpke here. In particular, you can get it as $\sigma$ of order 10, mapping $(5,6)$ to $(5,6)^\sigma = (1,2)(3,5)(4,6)$, leaving $(1,2,3,4,5)$ fixed, or as an automorphism of order 2, mapping $(1,2,3,4,5,6)\mapsto (2,6)(3,5,4)$ and $(1,2)\mapsto (1,2)(3,4)(5,6)$, etc. (If these don't agree with what you see, keep in mind that in GAP operations act on the right.)
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William DeMeoMar 25 '12 at 21:54

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@MichaelHardy: You want to look at the external semidirect product, which requires a group $N$, and a homomorphism $\varphi\colon H\to \mathrm{Aut}(N)$; then you define $N\rtimes_{\varphi} H$ as the set of pairs $(n,h)\in N\times H$ with product $$(n_1,h_1)(n_2,h_2) = (n_1\varphi(h)(n_2), h_1h_2).$$ With $H=\mathrm{Aut}(N)$, you get a natural action (via $\varphi=\mathrm{id}$), which creates a group $N\rtimes \mathrm{Aut}(N)$ known as the holomorph of $N$.
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Arturo MagidinMar 26 '12 at 0:00

2 Answers
2

As others have pointed out, you can embed any group in its holomorph. But for a group $G$ with trivial centre (like $S_6$), the normal subgroup ${\rm Inn}(G)$ of ${\rm Aut}(G)$ is naturally isomorphic to $G$, so you can embed $G$ directly into ${\rm Aut}(G)$, which is of course a smaller group than the holomorph.

In fact ${\rm Aut}(S_6)$ embeds into $S_{12}$ as follows. Let $\tau$ be an outer automorphism of ${\rm Aut}(S_6)$ with $\tau^2=1$. For an element $g \in S_6$ acting on the set $\{1,2,3,4,5,6\}$, let $\tau'(g)$ denote $\tau(g)$, but acting on the set $\{7,8,9,10,11,12\}$. So, for example, if $g=(1,2)$ and $\tau(g) = (1,2)(3,4)(5,6)$, then $\tau'(g) = (7,8)(9,10)(11,12)$.

Then we can embed $S_6$ into $S_{12}$ by $g \to g\ \tau'(g)$. Then $t := (1,7)(2,8)(3,9)(4,10)(5,11)(6,12)$ induces $\tau$ by conjugation on the image of this embedding, so $\tau$ together with this image generates ${\rm Aut}(S_6) \le S_{12}$.

This image is a subgroup of the Mathieu group $M_{12}$ and can be used as part of one of the many constructions of $M_{12}$.

More generally: if $N$ is a group, and $\varphi\colon H\to \mathrm{Aut}(N)$ is a group homomorphism, then this induces an action of $H$ on $N$ by the formula
$${}^hn = [\varphi(h)](n),$$
where "${}^hn$" means "the image of $n$ under the action of $h$". The formula makes sense because $\varphi(h)\in\mathrm{Aut}(N)$, so we can evaluate $\varphi(h)$ on $n$.

This action satisfies nice properties:

${}^h(nm) = {}^hn{}^hm$ for all $n,m\in N$, $h\in H$.

${}^k({}^hn) = {}^{kh}n$.

Given such a triple, $(N,H,\varphi)$, we can construct a $G$ that contains subgroups $\mathcal{N}$ and $\mathcal{H}$ isomorphic to $N$ and $H$, respectively, with $\mathcal{N}\triangleleft G$, $G=\langle\mathcal{N},\mathcal{H}\rangle=\mathcal{NH}$, and where for every $\mathfrak{h}\in \mathcal{H}$ and $\mathfrak{n}\in\mathcal{N}$, if $h\in H$ corresponds to $\mathfrak{h}$ and $n$ corresponds to $\mathfrak{n}$, then $\mathfrak{hnh}^{-1}$ corresponds to ${}^hn$. This is the semidirect product $N\rtimes_{\varphi}H$.

The underlying set of $N\rtimes_{\varphi}H$ is $N\times H$. The multiplication rule is
$$(n_1,h_1)\cdot (n_2,h_2) = \Bigl(n_1{}^hn_2,h_1h_2\Bigr).$$
I'll leave it to you to verify this group satisfies the properties given above, with

Conversely, if $G$ is a group, $N\triangleleft G$, $H\leq G$ are subgroups with $NH=\langle N,H\rangle = G$ and $N\cap H=\{1\}$, then for each $h\in H$ we have an automorphism of $N$ given by $n\mapsto hnh^{-1}={}^hn$; this yields a group homomorphism $\varphi\colon H\to \mathrm{Aut}(N)$, and it is easy to verify that the group $N\rtimes_{\varphi}H$ constructed as above is isomorphic to $G$.

Now, by taking $\varphi\colon\mathrm{Aut}(G)\to\mathrm{Aut}(G)$ being the identity, we can always construct the group $G\rtimes_{\mathrm{id}} \mathrm{Aut}(G)$, which is the aforementioned holomorph of $G$.